CJ1965 4-20-2018
Wrong again. You cannot simply substitute the electrical power ratio into the sound power formula and call it a day.
The ratio of measured sound power to reference sound power is 1/2 - not 1/4.
I posted the following in this thread, into which this debate had spilled over: Almarg 4-21-2018
Regarding the debate about the relation between SPL and speaker input
power, I found the following sub-page at the site which provided the
calculators that were referred to earlier:
http://www.sengpielaudio.com/calculator-efficiency.htm
Entering
various parameters into either of the two calculators closest to the
bottom of that page (one entitled "Sound Pressure Level and Amplifier
Power" and the other in the section entitled "Electro-Acoustic
Sensitivity") clearly confirms what Atmasphere, Erik_Squires and I have
all said on this subject. Namely that if the input power to a speaker is
changed by a given number of db, SPL at a given listening distance will
change by the same number of db. (As previously stated, this of course
assumes that the speaker is not being driven hard enough to cause
thermal compression in the drivers to become significant).
I also
found the following writeup at PSB’s site, which provides additional
confirmation. About 2/3 of the way down the page a table is provided
showing power vs. volume for an unnamed 87 db speaker used as an
example. Note that 40 watts results in a volume of 103 db, while a 6 db
reduction in that power level (to 10 watts) results in a volume that is 6
db less (97 db). While a 3 db change in power, from 1 watt to 2 watts,
changes the volume by 3 db, from 87 db to 90 db. And a 20 db change in
power, from 1 watt to 100 watts, changes the volume by 20 db, from 87 db
to 107 db.
http://www.psbspeakers.com/articles/Guide-to-Speaker-Specifications
Regards, -- Al |
Life is always interesting. Here again we have the one who is most insulting - to be the most sensitive... aw , darn shame ain’t it ...
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Sorry, wrong is wrong. And with that, I have to move on. I have better things to do with my time.
And with that, I’m done giving out free lessons in first year electrical engineering. Find someone else to insult.
Well I must say some days a much more interesting here than others. :) E |
Al - Yes indeed. These are all just rewrite. Since power is proportional to the square of the voltage:
(0.5 x 0.5) = 0.25
Now, calculating power difference:
10 log (0.25) = -6 dB - erik_squires
Wrong again. You cannot simply substitute the electrical power ratio into the sound power formula and call it a day. The ratio of measured sound power to reference sound power is 1/2 - not 1/4. And if you multiply the area factor in the formula for sound power in a medium, the ratio goes back to 1 - yielding 0 db. And with that, I’m done giving out free lessons in first year electrical engineering. Find someone else to insult. |
I get particularly displeased when a person is graciously made aware of a shortcoming— but carries blithely on.
I was taught that’s a true true sign of real ignorance.
|
Cj1965 has made mistakes. Just shows he’s no different than anyone else when it comes to knowledgeable discussion.
Here’s hoping he’s at least bright enough-from an EQ perspective- to realize he wouldn’t enjoy belittling, condescending, or harshly critical comments directed as a personal attack on him. If he is “bright” enough he will reflect on his “petty” attempts at personal disparagement and get on with his sincere apology. Man up. If he doesn’t apologize he’ll be remembered (forgotten) as an inconsiderate, tiny minded( as he just couldn’t see the big picture) cretin, who could do a bit of math-but not always correctly :( |
Al - Yes indeed. These are all just rewrite. Since power is proportional to the square of the voltage:
(0.5 x 0.5) = 0.25
Now, calculating power difference:
10 log (0.25) = -6 dB
Calculating voltage difference:
20 log (0.5) = -6 dB
In either case, the SPL at a reference distance, measured in dB, changes in proportion to the power OR voltage when either is expressed as dB assuming there is no compression in the driver.
Said another way, for a single driver:
Delta V dB = Delta W dB = Delta SPL dB
That's what's so cool about dBs!
Best,
E |
@cj1965
Where in my post did I mention power? I mentioned the relationship between output dB and input voltage.
Having said that, as @almarg has alluded, both formulas are true. Much like
P = V x I
and
P = (V x V) / R
They are mere re-writing of each other. Again, I encourage you to grab XSim to validate any formulas. It is pretty accurate.
Best,
E |
Assuming that a speaker is operating in a reasonably linear manner, meaning for example that it is not being over-driven to the point that thermal compression becomes significant, it seems to me that the relation between acoustic power out and electrical power in will remain constant to a close approximation. And electrical power in will be proportional to the square of the applied voltage.
Therefore it would seem to me (and I believe also to Erik, Atmasphere, and Kijanki) that since a 50% reduction in applied voltage will result in a 75% reduction in electrical power in, which corresponds to a 6 db reduction in electrical power in, the result will be a 6 db reduction in acoustic power out.
Regards,
-- Al
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@almarg
Please look up the definition of sound power. You will find it is not based solely on the net force applied but the area as well. If you run the calculations for sound power and put the values in the sound power db formula, you will get -3db. The graph below the calculator on the website Erik linked to also indicates this (voltage gain/loss is-6db while, sound power is -3db).
The formula or calculator for voltage gain/loss is not supposed to be used for sound power. One has force units the other power - apples and oranges.
.
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Thanks @almarg - I should have said "voltage" a few more times I think.
Best,
E |
... not 20 X log 10(P/Po) db as you [Erik] suggested above. CJ1965, Erik did not say that or suggest that. He said 20 x the logarithm of the ratio of applied voltage, not 20 x the logarithm of the ratio of power. If half the voltage is applied power will be reduced to 1/4 of the original amount, resulting in a 6 db reduction in SPL according to your own 10log(P/Po) formula, which is correct. And that is what Kijanki was implicitly pointing out when he quoted your erroneous original comment on the matter. Also, and more significantly, +1 to the very well said comments above by Peter (Ptss), which together with the earlier comments by Timlub I suggest that you (CJ1965) would be well advised to take to heart. Regards, -- Al |
@erik_squires No, we're not far off from the essence of the question posed by the OP. Application of the decibel expression of voltage gain/loss to sound power gain/loss is not appropriate..
Sound power level, denoted LW and measured in dB, is defined by
LW
=
10
log
10
(
P/P
0
)
d
B
not 20 X log 10(P/Po) db as you suggested above.
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@cj1965 and other Respected and Valued contributors to this forum. cj1965 is obviously well schooled in the math, a stickler for precision, and interested in the truth of the matter. I believe/hope his knowledge “could” become valued here. However, “manners” are vital in order to become a “respected and Valued” member of any reasonable group. “Anyone” can have a misstep or slip up. We’re human. Such an event “does not” necessarily diminish one’s value or credibility. I’ll mention Ralph here as he is known to produce exceptionally listenable and prized audio equipment under the ATMASPHERE label; and is without doubt a valued contributor.
However, harsh criticism, or sarcasm, is “generally recognized” as a deliberate attempt to demean a person.
Reasonable and compassionate people will/may react to defend the person(s) subject to the personal attacks.
This has occurred here.
I don’t support a ban of the offender at this time. He may have been having a “bad day(s) for reasons unknown. I do believe apologies are “generally accepted”; and are a good response to a correction of a error in a “statements of fact”, even if semantics are causing the misunderstanding, as well as for “personal insults”. Unfortunately, insults are more unacceptable than factual errors(which can always be corrected!). Insults, even if based on accurate facts, cannot be corrected. They linger in the minds of all. I highly recommend the high road of, at the minimum, a general apology for insulting comments. And best wishes to all. Peter
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Also, we are so far from the OP's original topic, maybe we should let this thread just die. :) |
For a single driver : The change in dB in input voltage is equal to the change in output SPL in dB. Voltage dB can be calculated in this fashion: dB = 20 x log ( Vnew / Vold ) in the case of half the voltage: 20 x log ( 1 / 2 ) ~= (- 6 dB) Also, you can use this handy web gizmo: http://www.sengpielaudio.com/calculator-gainloss.htm |
This is exact quote from cj1965 post:
In theory, half the voltage applied to a driver will result in half the sound pressure in the output (-3db). Res ipsa loquitur, huh? Obviously he didn't actually do the math, but based this on belief instead- the exact thing of which he was accusing others... |
Thanks, Kijanki. Here's another interesting quote, this one from the thread Tim referred to above:
CJ1965 4-19-2018
For some people, "results" means using antiquated 80 year old technology
that is highly vulnerable to changes in performance depending on what
it's connected to. And for some folks, the lack of bass and exaggerated
highs, coupled with increased harmonic distortion when using this
ancient "technology" is "pleasing" or "desirable". Similarly, others see
the pops and ticks, rapid wear, uneven high frequency performance,
limited dynamic range, increased distortion, wow, and flutter associated
with ancient vinyl technology as "more authentic". Unfortunately, we
can't confine such individuals who promote and buy this junk to padded
cells. We pretty much have to create invisible "padded cells" that
effectively allow ourselves to ignore them.
Best regards, -- Al |
This is exact quote from cj1965 post: In theory, half the voltage applied to a driver will result in half the sound pressure in the output (-3db). |
This was posted on another thread today. I don't think that he is 100% wrong, but the problem of his delivery does not seem to sink in. Its everyone else.
@cj165 ... with my pea brain, I had to look up fiefdom syndrome. Thanks for the education there. I picked a few things from your rant for you to consider.
1. You have not been scorned or ridiculed because you challenged the resident experts.... You are correct, you are scorned, but it is because of your disrespect.... Overall, I loved the debate, Your contribution will cause many people to start studying to get to the last nitty gritty that separated your opinions in the debate.
2. Others "Command Respect because of their post count" Their is no Command... I assume that you are referring to Al & Ralph. They both regularly contribute in a way that has helped countless people solve technical issues in their own systems.
3. "Followers implore moderators to ban the newbie" That could get worse, its up to you. I don't believe that anyone would want you to change your opinion and sharing facts. Its all in the presentation.
4. "Causing issues to constantly being revisited" Yes, issues are often revisited, but it is because, they come up every so often for different people and are often addressed as they come. It has nothing to do with Impeding Technical progress.
Everyone, Here is his post from this morning, See Below:
"
Some Audiogoners (no names) start threads after threads which ask the
same questions, over and over and over again. They usually get
thoughtful responses, but they, the OPs, don’t seem to learn anything or
have the motivation to act on any of the thoughtful responses. At best,
"they" seem confused or maybe have too much time on their idle hands.
" - ps
In Audiogon forums, the ratio of the clueless soothing their fragile egos with endless hand waiving nonsense to the knowledgeable citing proven science and mathematical evidence tends to be pretty high. This scenario is clearly not restricted to Audiogon alone. A high percentage of forums possess posters who "command respect from followers" simply because they have several thousand posts under their belt. The fact that most of their contributions have been hand waiving BS intentionally or unintentionally misleading others doesn't seem to matter. The "reputation" and "belief system" are all that counts. And when a newcomer or "unknown" hits the scene with "facts from the big bad world outside", he/she is often scorned, ridiculed, challenged for credentials, or in egregious cases - the "followers" implore moderators to ban the newbie if the newbie presents facts that disagree with the "grand proclamations of resident forum experts". The end result is often persistent ignorance, in many cases, this ignorance persists with basic fundamental subjects that "in the big bad world outside", have been settled long ago and are considered trivial. This fiefdom syndrome is made worse when utterly unqualified salesmen are allowed to pitch their products without any deference to science or facts. The mere fact that they've been pitching the same old garbage for 30 years is supposed to be enough for the "followers" and sadly in many cases, it is. So just as it is with other forums, at Audiogon, there are a significant number of influences that impede technical progress - causing issues to be constantly revisited that should have been settled long ago. |
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@cj165
I'm sorry, I would call you by name, but I haven't seen it posted. I came very close to sending this in a private message, but much has been addressed in this thread as well as a couple of others.
1st... You are extremely knowledgeable and could be if you chose to be, a very strong contributing member to this forum. I personally would love to see that happen.
You have 7 pages of responses, I did not count the threads, but I would guess maybe 8 or 10. In those threads I read maybe 20 or so of the post. In each thread at some point, you belittled someone.
Odd that one of your threads, you discussed heavily forum etiquette, I believe the thread was discussing if someone needed to prove that they were qualified to chime in on these forums.
I would say that 98 percent of the members here come here to learn and share the passion of their hobby. All of us at some point get challenged with our knowledge and being the he men that we are, we seem to have to defend ourselves (except maybe Elizabeth and you attacked her also). I'm not sure if you are trying to prove something or are insecure in yourself that you have to feel superior over others? Really, I'm not try form a conjecture. I am really hoping that you are here wanting to contribute and would take it serious that we are all people, all have feelings and all want to be treated with respect. There is no doubt, that I have said stupid things on this forum that can and have been challenged, On the other hand, there is no doubt, that I have made a difference and contributed to others personal audio issues. Your knowledge in most area's of audio make my knowledge a pittance, but I can tell you if the disrespect continues, I will turn you off. I appreciate your knowledge level and am hoping to appreciate you personally for your contributions.
I am trying to say all of this in respect.
Tim (timlub)
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Each is not absorbing 1/2 watt when placed in series (with the required
2.83 V applied) and the net acoustical sum of output is approximately
97db - not 94db. Two obvious errors in one sentence. The math doesn't
lie and frankly, I'm getting a little tired of the verbal gymnastics at
this point. The math is simple and anyone who wants to read the details
above can readily see where the error was established. Yeah, you can
double the input drive voltage to 5.7v and then the series arrangement
will have each woofer dissipating 1/2 watt. You can also multiply the
input voltage by 10,000 - it won't tell us anything about the effect
placing two identical drivers in series has. @almarg Its pretty evident to me that this guy seems to equate '1 watt' with '2.83v' and so is somehow assuming that I meant '2.83 volts' when I was pretty talking about the power (not voltage) being '1 watt'. Also his intuitive grasp of Kirchoff's Law seems to be weak, and I further find that talking about capacitance when dealing with 2 guitar speakers as opposed to one rather silly, since any such distributed capacitance on the coils (or anywhere else) is going to be quite low and negligible. But even if it was not, in a series connection current would have to flow through both drivers in order for one to work. For all his 'hand waving', he seems to have no grasp of the formulae he presented, how they are used or their significance. @kijanki +1 |
As quoted by CJ1965: " So if you have a 97 db 1 watt/1 meter 8 ohm driver, two in series will have the same efficiency (since each is absorbing 1/2 watt) while the sensitivity is 94 db " - atmashpere [sic] Given that efficiency is defined as the SPL produced at 1 meter in response to 1 watt, when he said "since each is absorbing 1/2 watt" he clearly meant "since each is absorbing 1/2 watt if a total of 1 watt is being provided." Also, the paragraph in which that statement appeared certainly made no reference to 2.83 volts. If that was not clear then it certainly should have become clear during the course of the subsequent posts, especially mine. ... you can double the input drive voltage to 5.7v and then the series arrangement will have each woofer dissipating 1/2 watt. You can also multiply the input voltage by 10,000 - it won’t tell us anything about the effect placing two identical drivers in series has. Specifying the efficiency of a series combination of speakers, defining efficiency as the SPL produced at 1 meter in response to 1 watt, will tell us whatever can be told by an efficiency spec. As Ralph alluded to earlier, efficiency specs tend to be especially relevant in the case of tube amps, since for example in the case of an amp providing 4 and 8 ohm output taps maximum power ratings (in watts) will typically be the same or similar when an 8 ohm load is connected to the 8 ohm tap as when a 4 ohm load is connected to the 4 ohm tap. Regards, -- Al |
"
So if you have a 97 db 1 watt/1 meter 8 ohm driver, two in series will
have the same efficiency (since each is absorbing 1/2 watt) while the
sensitivity is 94 db
" - atmashpere
Each is not absorbing 1/2 watt when placed in series (with the required 2.83 V applied) and the net acoustical sum of output is approximately 97db - not 94db. Two obvious errors in one sentence. The math doesn't lie and frankly, I'm getting a little tired of the verbal gymnastics at this point. The math is simple and anyone who wants to read the details above can readily see where the error was established. Yeah, you can double the input drive voltage to 5.7v and then the series arrangement will have each woofer dissipating 1/2 watt. You can also multiply the input voltage by 10,000 - it won't tell us anything about the effect placing two identical drivers in series has. Sorry, wrong is wrong. And with that, I have to move on. I have better things to do with my time.
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CJ1965 4-18-2018
...
the erroneous statement atmasphere made and you [Almarg] supported which
suggested two 8 ohm speakers in series would each dissipate 1/2 watt
with a total of 2.83V applied as input.
Neither I nor Ralph (Atmasphere) said or even implied that.
You can't apply or "provide a watt". All you can do is apply a voltage
and the load draws whatever current it draws based on its resistance.
What we were referring to is applying whatever voltage is necessary to result in a total of 1 watt being consumed by the two speakers, whether they are connected in series or in parallel. Which is what is relevant if what is being referred to is the overall efficiency of the speaker combination, and if efficiency is defined as the SPL produced at 1 meter in response to 1 watt. I'm not sure how that can not be clear, after all that has been said. Regards, -- Al
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I'm voting to remove cj1965 from our forum. Administrator should already observe that he calls respected members "clueless". I wonder who would join me and what is the procedure to remove such obstacle. |
" re your last post that was directed to me, your math is of
course correct. However, once again, if efficiency is defined as the SPL
produced at 1 meter in response to an input of 1 watt, if that 1 watt
is provided to two speakers connected either in series or in parallel
each speaker will absorb half of a watt.
" - almarg
Please, I cannot take you seriously if you (or anyone else for that matter) continues saying things like - "if that one watt is provided" to loudspeakers. You can't apply or "provide a watt". All you can do is apply a voltage and the load draws whatever current it draws based on its resistance. The accuracy of my math above stands and it stands in stark contrast to the erroneous statement atmasphere made and you supported which suggested two 8 ohm speakers in series would each dissipate 1/2 watt with a total of 2.83V applied as input. Wrong is wrong. Math is either correct or incorrect. 1/2 watt DOES NOT equal 1/4 watt. Period.
.
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"
For someone who has only been participating in this forum for a couple
of months you certainly are quick to direct insulting comments at some
of the forum’s most knowledgeable and widely respected members. And that
certainly includes Kijanki, as well as Atmasphere. I suggest that some
modification to your manner would result in your contributions being
better received, and discussions in which you participate being more
constructive.
" - almarg
Perhaps you should direct your comments to the folks with vast, impeccable credentials who say things like this:
"
You're telling Almarg to do the math? LOL, Man, you got big mouth.
" - kijanki
Btw, I know my math is correct, thank you. I should hope after four years of electrical engineering, I would know a little about basic addition, subtraction, multiplication, and division....
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" Really? Hand-waving seems to be what you are doing with this quote...
Somehow you don’t seem to make the connection that current has to flow if a circuit is complete. Its Circuit Basics 101 first day stuff. " - atmasphere
What a clueless statement. When a capacitance is present, current flow is dependent on the time rate of change of voltage across the capacitor. Since it is a series circuit, this time rate of change affects the flow of current with time through both inductors - Circuit Basics 102.
I= Cdv/dt
Your simplistic view assumes that current through all reactive elements in a circuit is steady state (constant). And that is pretty scary for someone who has experience building amps. But then again, it’s tube amps and this is the year 2018 - so that kinda makes sense.
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CJ1965, re your last post that was directed to me, your math is of course correct. However, once again, if efficiency is defined as the SPL produced at 1 meter in response to an input of 1 watt, if that 1 watt is provided to two speakers connected either in series or in parallel each speaker will absorb half of a watt. So when efficiency is referred to, and defined as the SPL produced at 1 meter in response to 1 watt, the results of your calculation need to be normalized to 1 watt of supplied power. Which in turn makes Ralph’s statement that you alleged to have "blown it" entirely correct. You [Kijanki] could know something that no one else on the planet knows and we’re all in for a new, amazing discovery. But then again, you just might be another clueless individual trolling around on the internet in search of a pointless argument.... For someone who has only been participating in this forum for a couple of months you certainly are quick to direct insulting comments at some of the forum’s most knowledgeable and widely respected members. And that certainly includes Kijanki, as well as Atmasphere. I suggest that some modification to your manner would result in your contributions being better received, and discussions in which you participate being more constructive. Regards, -- Al |
I'm the only one in this thread actually citing equations based on
Kirchoff's law while you hand wave generalizations about tube amps. Really? Hand-waving seems to be what you are doing with this quote... Somehow you don't seem to make the connection that current has to flow if a circuit is complete. Its Circuit Basics 101 first day stuff. |
@atmasphere
I'm the only one in this thread actually citing equations based on Kirchoff's law while you hand wave generalizations about tube amps.
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This jives with measurements I've made of acoustic output of drivers in series. Sounds like you are running a solid state amp. Solid state amps, generally speaking, are usually built to behave as a voltage source. If that is the case, it will make double the power into half the impedance, and thus also half the power if the impedance is doubled. So such an amplifier will make 1/4 of its 4 ohm power into 16 ohms (drivers in series). |
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I'm giving up - you won't get it.
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This bit is problematic. The drivers are going to sound the same whether in parallel or in series. What **won’t** sound the same is the amplifier, which reacts very differently to impedance depending on the amp.
Hi Ralph,
I believe that I did elude to the fact that it is the amp that sound differently, but I want to stress that in this case, compensation circuitry should sound better with tubes that the speaker without the circuitry.
Using 2 - 16 ohm speakers, depending on frequency, you will see impedances normally rise to 30 or 40 ohms if not higher. You can obviously go into why and amp sounds better on 8 ohms than 4. Impedance compensation circuitry normally only lowers the peaks (stated for others on the thread), it never raises an impedance dip. So when we can make a speaker consistent from say 8 to 9 ohms, in my experience A tubed amplifier putting out consistent power across all frequencies has always sounded better than an amp that sees a speaker that strays radically. Again, this is only my experience, today, I seldom build a speaker without compensation, there are a few exceptions, but Few.
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"Again, current (and not voltage) is what moves the coil." - kijanki
That statement is utterly clueless and contradicts a widely held definition that has been born out in countless measurements of voltage and current in coils. Once again, the definition of voltage as it RELATES to current in any given inductor coil is:
V(t)= Ldi/dt
where voltage across an inductor at any given time is directly proportional to the time rate of change of current running through said inductor. Voltage and current in an inductor are INEXTRICABLY LINKED to one another and the precise mathematical representation has been given earlier in this thread. If you disagree with the above equation, good for you. You might be on to a new discovery in electronics and physics. Somehow, I doubt it. You could know something that no one else on the planet knows and we're all in for a new, amazing discovery. But then again, you just might be another clueless individual trolling around on the internet in search of a pointless argument....
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Again, current (and not voltage) is what moves the coil. |
Incorrect? Really? In theory, they move at EXACTLY the same time only
when no capacitance exists in the circuit. Current through the coils is
the same at any given time only when the circuit doesn’t have
capacitance. We know this is not the case. @cj1965 It seems that you are forgetting about Kirchoff's Law. Obviously one woofer cannot move if current is not also flowing through the second in a series connection! So that means the other woofer has to move **at the same time** otherwise Kirchoff's Law is violated, which is an impossibility. Further, the current has to also flow the other way (it is an audio waveform after all), so the situation with the woofer's relationship is reversed; it is quite obvious that they will move at exactly the same time and this is true even if the drivers are of different inductance. Kirhoff's Law was also taught to me in school as the 'Law of energy conservation'. It basically states that there cannot be more energy in a circuit than is put into it, nor can there be any less. So if a watt is put in, the individual parts in the circuit will all dissipate some fraction of that watt in such a way that if you added it all up, it would be exactly 1 watt. Here is the Wikipedia page: https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_lawsNow if the drivers are dissimilar, they might not *sound* the way you want (one might have a greater voltage drop across it than the other), but if they are the same driver there isn't a downside unless higher impedance is problematic as with a solid state amp (however the solid state amp will be seen to be making less distortion). If all things are equal, meaning a 4 ohm 8 ohm and 16 ohm driver would
all sound equivalent in a given cabinet, Paralleled wired speakers sound
better overall than series. I would also recommend that you add a
simple impedance correction circuit. Many amplifiers react better to a
consistent impedance. Not all, but No amplifier sounds worse with it.
(unless you are dropping impedance too low for a few tubes) still within
tubes impedance operating range, even tubes sound better with
impedance correction circuits on speakers
This bit is problematic. The drivers are going to sound the same whether in parallel or in series. What **won't** sound the same is the amplifier, which reacts very differently to impedance depending on the amp. Now in the case of this thread, the amp in question is a tube guitar amp. Tubes, generally speaking, prefer a higher impedance and will make less distortion if the higher impedance is accommodated by a tap on the output transformer. The transformer will run cooler as it is more efficient, so with most tube amps you get a tiny bit more power as well, as well as more extension into the bottom octave of the amp. Reducing distortion may not be the goal in a guitar amp, where the guitarist's individual 'sound' that he is going for is highly subjective and varies greatly from one guitarist to another, even using the amp amp. So ultimately, the OP will simply have to try it both ways. The amp won't be damaged by this, and if there are taps for each impedance, they should be employed. |
You're telling Almarg to do the math? LOL, Man, you got big mouth. |
"
@cj1965 Current in the circuit is still the same. You are confusing
voltage across the speaker with current. Imagine simple circuit
consisting of voltage source and bunch of inductors and capacitors in
series. Now you insert speaker into it. Do you think it will sound
differently in different places of insertion? There is only one current
in the circuit and two speakers in series have to respond at the same
time (unless there is place where "faster current" can escape)." - kijanki
Please read and study the equation I posted above for two inductors (speaker coils) in series with a resistance and capacitance. The equation doesn't lie. Current and voltage are constantly varying and as the equation shows the voltage representations of each woofer are NOT equivalent to the applied votlage input [V(t)] ,minus the other woofer's voltage. The voltage represented by the capacitance must also be accounted for and it is time dependent. If you disagree with the equation and what it is saying - address your comments/concerns with that equation and its applicability to the subject at hand. I didn't invent Kirchoff's law - I'm merely reciting it in the context of a series connected loudspeaker pair.
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@almarg
Do the math. Take two 8 ohm nominal woofers, connect them in series, and apply 2.83v.
2.83 divided by 16 = .176875 Amps (current "I")
power dissipated into each woofer in this circuit then becomes:
R(I X I) or resistance of each woofer times the square of the current running through it. Thus
8(.176875 X .176875) = .25 Watts or 1/4 watt - not 1/2 watt as atmasphere and you have suggested -
"
two in series will have the same efficiency (since each is absorbing 1/2 watt) while the sensitivity is 94 db
"- atmasphere
The above can be expressed another way.
Power dissipated in two woofers in series = net voltage drop across both times the net current flowing through both.
2.83 Volts times .176875 = .5 watts or 1/2 watt total between both drivers in series. If together both woofers are dissipating 1/2 watt, then individually they must dissipate 1/4 watt - thus agreeing with the above calculation. At half the voltage drive level (voltage divided when connected in series), each woofer should produce half the acoustic output and consume 1/4 the power.When their acoustic outputs are summed, the net acoustic result is the same as if for one driver - however the power consumed is cut in half.
This jives with measurements I've made of acoustic output of drivers in series.
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@cj1965 Current in the circuit is still the same. You are confusing voltage across the speaker with current. Imagine simple circuit consisting of voltage source and bunch of inductors and capacitors in series. Now you insert speaker into it. Do you think it will sound differently in different places of insertion? There is only one current in the circuit and two speakers in series have to respond at the same time (unless there is place where "faster current" can escape).
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CJ1965 4-17-2018
"So if you have a 97 db 1 watt/1 meter 8 ohm driver, two in series will have the same efficiency (since each is absorbing 1/2 watt) while the sensitivity is 94 db. If you put the two drivers in parallel for a 4 ohm load, the efficiency is the same as 1/2 is absorbed by each driver if 1 watt is applied. However the sensitivity is now 100db. " - atmasphere
With that, you clearly blew it. Ralph’s (Atmasphere’s) statement is entirely correct, if (as he is assuming, consistently with commonly seen usage) efficiency is defined on a 1 watt basis. And if sensitivity is defined on the basis of 2.83 volts. Strictly speaking, I would define speaker efficiency as being the ratio of acoustic power out to electrical power in. But that definition would have little practical use, and the term is widely (and IMO very reasonably) used to refer to the SPL produced at 1 meter in response to an input of 1 watt. Regards, -- Al |
CJ1965, I believe that what your analysis about phase differences between voltage and current in reactive circuit elements may not be taking into account is that the motion of a dynamic driver results from the current it is conducting, rather than from the voltage that is applied to it. And the current in a series circuit is of course identical at all points in the circuit, aside from the consequences of the propagation delay that will exist along that path. Which in turn will be completely negligible for path lengths that would be present in a home environment, putting aside reflection effects for which timing may have relevance in digital applications.
Regards,
-- Al
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Wow, I am soooo glad that I didn't spot this thread before now.....
giving way to all the theory and mathematics,
@williewonka
If all things are equal, meaning a 4 ohm 8 ohm and 16 ohm driver would all sound equivalent in a given cabinet, Paralleled wired speakers sound better overall than series. I would also recommend that you add a simple impedance correction circuit. Many amplifiers react better to a consistent impedance. Not all, but No amplifier sounds worse with it. (unless you are dropping impedance too low for a few tubes) still within tubes impedance operating range, even tubes sound better with impedance correction circuits on speakers.
NO PROOF, this is my experience.
Good Luck, Tim
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"
Thank you. People need to learn to express their opinions without the
need to insult others, especially the ones whose knowledge and
experience as well as their contributions are unique in this industry.
" - kalali
If I had known that atmasphere was some "world class" amplifier designer, I probably would have used a different circuit analogy to make my point. In this case, I didn't have a clue who he/she was which re enforces a point I've been making elsewhere - when it comes to sharing accurate information with others, the credentials of the person attempting to share information are irrelevant . Only the facts and accuracy of the information matter.
Secondly, when you are a professional in an industry and put yourself out on a forum sharing "facts" to help educate others, you should accept the possibility that you will be "incorrect" or "wrong" on occasion and all of the downside associated with that potential. This underscores the need to be careful, exercise restraint, and double check your "advice" before you give it out. The outcome can solidify your reputation for expertise or weaken it - that's the inherent risk in sharing with others.
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+1 Al
Thank you. People need to learn to express their opinions without the need to insult others, especially the ones whose knowledge and experience as well as their contributions are unique in this industry.
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